

A248705


The cubes related to the strictly increasing subsequence of A053668(n), n >= 1.


1



1, 8, 27, 64, 343, 729, 2744, 3375, 6859, 35937, 46656, 148877, 287496, 438976, 778688, 2985984, 3869893, 8489664, 34645976, 43986977, 58863869, 75686967, 398688256, 426957777, 485587656, 596947688, 835896888, 1693669888, 2548895896, 2954987875, 4758586568
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OFFSET

1,2


COMMENTS

The triangular numbers of this form are at A246753.
The squares of this form are at A248648.


LINKS



EXAMPLE

a(4) = 64 = 4*4*4, which is a cube. Product of its digits = 6*4 = 24.
a(5) = 343 = 7*7*7, which is a cube. Product of its digits = 3*4*3 = 36.
Since 36 > 24, 64 and 343 appear in the sequence.
(Start)
A053668 is sieved (from left to right):
1, 2, 3, 4, 5, 6, 7, 8, 9, ....(numbers: k)
1, 8, 27, 64, 125, 216, 343, 512, 729, ....(cubes: k^3)
1, 8, 14, 24, 10, 12, 36, 10, 126, ....(prod of digits of k^3)
1, 8, 14, 24, X, X, 36, X, 126, ....(sieved products)
and related leftover cubes are:
1, 8, 27, 64, 343, 729, ....(leftover cubes)
(End)


MATHEMATICA

A248705 = {}; t = 0; Do[s = Apply[Times, IntegerDigits[n^3]]; If[s > t, t = s; AppendTo[A248705, n^3]], {n, 1, 10^4}]; A248705


PROG

(PARI) \\ For bfile
c = 0; k = 0; for(n=1, 5*10^8, d = digits(n^3); p = prod(i = 1, #d, d[i]); while(p > k, c++; print(c, " ", n^3); k = p))
(Python)
from operator import mul
from functools import reduce
A248705_list, x, m = [], 0, [6, 6, 1, 0]
for _ in range(10**9):
for i in range(3):
m[i+1]+= m[i]
xn = reduce(mul, [int(d) for d in str(m[1])], 1)
if xn > x:
x = xn


CROSSREFS



KEYWORD

nonn,base


AUTHOR



STATUS

approved



